Answer:
40 N
Explanation:
calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.
1 answer:
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The magnitude of electric field is produced by the electrons at a certain distance.
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C